2
$\begingroup$

I just want to clarify one thing I was never really sure on.

First the question:

$$1 = \lim_{x \rightarrow 0} \frac{e^x-1}{x} = \frac{\lim_{x \rightarrow 0} (e^x) -1}{\lim_{x \rightarrow 0} x}$$

is this valid?

I ask for two reasons:

1) because there are several rules that would make it seem wrong to do

2) If this is technically valid I can replace $\lim_{x \rightarrow 0} (e^x)$ with $\lim_{x \rightarrow 0} (1+x)$ (see [1] for explanation)

First

$$\lim_{x \rightarrow a} \frac{f(x)}{g(x)} = \frac{\lim_{x \rightarrow a} f(x)}{\lim_{x \rightarrow a} g(x)} \iff \lim_{x \rightarrow a} g(x) \ne 0$$

and

$$\lim_{x \rightarrow a} f(x) = f(\lim_{x \rightarrow a} x) \iff \lim_{x \rightarrow a} f(x) = f(a)$$

The second doesn't seem to be as much of an issue and I know it looks like I am answering my own question, but I am genuinely unsure of there is anyway to go about this without being forced to assume the limit (I don't want to use L'Hopital's rule, as it results in circular logic, neither proving or disproving the limit).

[1] $$\begin{align*}e &= \lim_{x \rightarrow 0} e = \lim_{x \rightarrow 0} (1+x)^{1/x} \\ &\Rightarrow (\lim_{x \rightarrow 0} e)^{\lim_{x \rightarrow 0} x} = (\lim_{x \rightarrow 0} (1+x)^{1/x})^{\lim_{x \rightarrow 0} x} \\ &\Rightarrow \lim_{x \rightarrow 0}(e^x) = \lim_{x \rightarrow 0} (1+x)^{\lim_{x \rightarrow 0} {1/x} * \lim_{x \rightarrow 0} x} \\ &= \lim_{x \rightarrow 0} (1+x)^{\lim_{x \rightarrow 0} {x/x}} = \lim_{x \rightarrow 0} (1+x) \\ &\therefore \lim_{x \rightarrow 0}(e^x) = \lim_{x \rightarrow 0} (1+x) \end{align*}$$

$\endgroup$
  • $\begingroup$ In terms of [1]], I can say that you can use maclauren series representation of exp(x) = 1+x+0.5x^2+ ... $\endgroup$ – Cardinal Jun 16 '16 at 20:08
  • 4
    $\begingroup$ Of course it's not valid, since you get the undefined expression “0/0”... $\endgroup$ – Hans Lundmark Jun 16 '16 at 20:09
  • $\begingroup$ @Cardinal, I don't want to do it that way since to build that series you must either 1) assume that exp(x)=1+x+1/2x^2+... or 2) use derivatives to build it, and both of which defeat the point of this exercise. $\endgroup$ – CopaceticMan Jun 16 '16 at 20:25
3
$\begingroup$

Your approach to evaluate the limit $\lim\limits_{x \to 0}\dfrac{e^{x} - 1}{x}$ fails because you are trying to use quotient rule for limits which works only when the limit of denominator is non-zero.

The justification given in $[1]$ is completely wrong although it does look like it is intuitively correct. The result $$\lim_{x \to 0}e^{x} = \lim_{x \to 0} 1 + x$$ is correct however because both the limits are equal to $1$. You need to think of limits as an essentially non-algebraic operation and you need to use the rules of limits when dealing with them. The justification in $[1]$ does not seem to use any of the standard rules of limits and seems more like an algebraic manipulation of the intuitive identity $e = (1 + x)^{1/x}$ when $x \to 0$.

Please don't do manipulations like these on limits. I will show the kind of problem which happens when you use continue to use such manipulation further. Let's start with the last equation of your justification $[1]$. \begin{align} \lim_{x \to 0}e^{x} &= \lim_{x \to 0}1 + x\notag\\ \text{and }\lim_{x \to 0}1 + x &= \lim_{x \to 0} 1 + 2x\notag\\ \Rightarrow \lim_{x \to 0}e^{x} &= \lim_{x \to 0} 1 + 2x\notag\\ \Rightarrow \lim_{x \to 0}\{e^{x}\}^{1/x} &= \lim_{x \to 0} (1 + 2x)^{1/x}\notag\\ \Rightarrow e &= e^{2}\notag \end{align} The above justification like yours is completely wrong. If using above justfication you replace $\lim_{x \to 0}e^{x}$ by $\lim_{x \to 0}1 + 2x$ you will get $\lim\limits_{x \to 0}\dfrac{e^{x} - 1}{x} = 2$ by your method.

The only justified way to evaluate the limit $$\lim_{x \to 0}\frac{e^{x} - 1}{x}\tag{1}$$ is to use a suitable definition of symbol $e^{x}$ and then proceed further. One such approach is given here. Another simpler approach is to consider $\log x$ defined by $$\log x = \int_{1}^{x}\frac{dt}{t}\tag{2}$$ and consider $e^{x}$ as inverse function to logarithm so that $e^{x} = y \Leftrightarrow \log y = x$. With this approach we can see from $(2)$ that $$\frac{d}{dx}\{\log x\} = \frac{1}{x}$$ and hence derivative of $f(x) = \log x$ at $x = 1$ is $1$. This implies (using limit definition of derivative) that $$\lim_{h \to 0}\frac{\log(1 + h)}{h} = 1\tag{3}$$ and putting $\log(1 + h) = x$ we see that as $h \to 0$ we have $x \to 0$ and $h = e^{x} - 1$ so that $(3)$ is transformed into $$\lim_{x \to 0}\frac{x}{e^{x} - 1} = 1$$ or $$\lim_{x \to 0}\frac{e^{x} - 1}{x} = 1$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.